Finding $ \prod_{n=1}^{999}\sin\left(\frac{n \pi}{1999}\right)$ I would appreciate if somebody could help me with the following problem. How can we find the product 
$$ \prod_{n=1}^{999}\sin\left(\frac{n \pi}{1999}\right)$$
 A: I have an even more precise bound: the product is less than
$$(999)! \left(\frac{\pi}{1999}\right)^{999}$$
Use Stirling approximation to get that this in turn is less than
$$ \sqrt{999 e} \left ( \frac{999 \pi}{1999 e}\right )^{999} \approx 4 \times 10^{-237}$$
A: The first digits of your number are:
0.00

To get this estimate notice that all terms of the product are less than $1$ and that $|\sin(x)| \le |x|$.
To be more precise your number is between $0$ and $10^{-80}$.
A: $$\prod_{k=1}^{n-1}\sin\left(\frac{k\pi}{n}\right)=\frac{n}{2^{n-1}}$$

Here, let $$x=\prod_{k=1}^{999}\sin\left(\frac{k\pi}{1999}\right) \tag{1}$$
Since $\sin t=\sin (\pi-t)$, therefore,
$$x=\prod_{k=1}^{999}\sin\left(\frac{(1999-k)\pi}{1999}\right)=\prod_{k=1000}^{1998}\sin\left(\frac{k\pi}{1999}\right) \tag{2}$$
Multiplying equation $(1)$ by equation $(2)$ gives,
$$x^2=\prod_{k=1}^{1998}\sin\left(\frac{k\pi}{1999}\right)=\frac{1999}{2^{1998}}$$ $$\implies x=\frac{\sqrt{1999}}{2^{999}}$$
We took $x>0$ because all angles are in $1^{st}$ and $2^{nd}$ quadrant.
